HDU 3709 Balanced Number（数位DP）

A balanced number is a non-negative integer that can be balanced if a pivot is placed at some digit. More specifically, imagine each digit as a box with weight indicated by the digit. When a pivot is placed at some digit of the number, the distance from a digit to the pivot is the offset between it and the pivot. Then the torques of left part and right part can be calculated. It is balanced if they are the same. A balanced number must be balanced with the pivot at some of its digits. For example, 4139 is a balanced number with pivot fixed at 3. The torqueses are 4*2 + 1*1 = 9 and 9*1 = 9, for left part and right part, respectively. It's your job 
to calculate the number of balanced numbers in a given range x,yx,y.

Input

The input contains multiple test cases. The first line is the total number of cases T (0 < T ≤ 30). For each case, there are two integers separated by a space in a line, x and y. (0 ≤ x ≤ y ≤ 10 ¹⁸).

Output

For each case, print the number of balanced numbers in the range x,yx,y in a line.

Sample Input

2
0 9
7604 24324

Sample Output

10
897

Hint

给定区间[a,b]，求区间内平衡数的个数。所谓平衡数即有一位做平衡点，左右两边数字的力矩想等。

遍历每一位做为平衡点，进行搜索，sum保存数字乘以距离的和，若sum为0，则说明平衡。 
要注意因为遍历了len次，所以0多加了len－1次。 
还有个小技巧是当sum<0时就可以直接return了，可以加速。因为，len由大到小的过程中，sum是由大到小的变化，但绝不会小于0，否则就是不能平衡。

#include <bits/stdc++>

using namespace std;
long long dp[20][20][1540]; /// dp[i][j][k] i表示处理到的数位pos，j是支点，k是力矩和
int bit[20];

long long dfs(int pos,int pivot,int pre,bool flag)
{
    if(pos==-1)return pre==0;  ///出口，也就是正好左右平衡时
    if(pre<0)return 0;         ///当前力矩为负，剪枝
    if(!flag&&dp[pos][pivot][pre]!=-1)
        return dp[pos][pivot][pre];
        
        
    int end=flag?bit[pos]:9;
    long long ans=0;
    for(int i=0;i<=end;i++)
        ans+=dfs(pos-1,pivot,pre+i*(pos-pivot),flag&&i==end);  ///i*(pos-pivot)写的很巧妙，支点左边加，右边减，支点不算
        
        
    if(!flag)  dp[pos][pivot][pre]=ans;
    return ans;
}
long long calc(long long n)
{
    int len=0;
    while(n)
    {
        bit[len++]=n%10;
        n/=10;
    }
    long long ans=0;
    for(int i=0;i<len;i++)  ///遍历每一个中心点
        ans+=dfs(len-1,i,0,1);
    return ans-(len-1);///去掉全0的情况
}
int main()
{
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
    int T;
    long long x,y;
    memset(dp,-1,sizeof(dp));//这个初始化一定别忘记
    scanf("%d",&T);
    while(T--)
    {
        scanf("%I64d%I64d",&x,&y);
        printf("%I64d
",calc(y)-calc(x-1));
    }
    return 0;
}